Problem: Let $y=\sin(x)\cos(x)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\cos(x)+\sin(x)$ (Choice B) B $\cos^2(x)+\sin^2(x)$ (Choice C) C $\cos^2(x)-\sin^2(x)$ (Choice D) D $\cos(x)-\sin(x)$
Answer: $\sin(x)\cos(x)$ is the product of two, more basic, expressions: $\sin(x)$ and $\cos(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}(\sin(x)\cos(x)) \\\\ &=\dfrac{d}{dx}(\sin(x))\cos(x)+\sin(x)\dfrac{d}{dx}(\cos(x))&&\gray{\text{The product rule}} \\\\ &=\cos(x)\cdot \cos(x)+\sin(x)\cdot (-\sin(x))&&\gray{\text{Differentiate }\sin(x)\text{ and }\cos(x)} \\\\ &=\cos^2(x)-\sin^2(x)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\cos^2(x)-\sin^2(x)$